1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sequence of measurable subsets of [0,1] (Lebesgue measure, Measurable)

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\left\{E_{k}\right\}_{k\in N}[/itex] be a sequence of measurable subsets of [0,1] satisfying [itex]m\left(E_{k}\right)=1[/itex]. Then [itex]m\left(\bigcap^{\infty}_{k=1}E_{k}\right)=1[/itex].

    2. Relevant equations
    m denotes the Lebesgue measure.
    "Measurable" is short for Lebesgue-measurable.

    3. The attempt at a solution
    My intuition was to construct 2 disjoint and uncountable subsets of an uncountable set that all have measure equal to 1.

    After much struggling, I don't think this is possible. To be able to construct such a counterexample would mean there is a mapping of each subset to the natural numbers which would make them both countable.

    Are there non-constructive approaches to building this counterexample? My attempts haven't yielded anything meaningful. Each example (rational and irrational, transcendental and algebraic) I've considered ends up with measure 1 and measure 0 subsets instead of both measure 1.

    I am not sure if my counterexample building skills are lacking or I just need to adjust my intuition. Any help would be appreciated.
  2. jcsd
  3. May 29, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    What is the measure of the complement (in [0,1]) of ##\cap_{k=1}^{\infty} E_k##?
  4. May 29, 2013 #3
    If I can construct a counterexample of 2 disjoint and noncountable subsets on [0,1] of measure 1, then the measure of the complement of intersections would be 1. However, I have not been able to find such a counterexample. The answer based on what I have considered is 0.
  5. May 29, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    So, you do not intend to answer my question. Fair enough, but that's it for me.
  6. May 29, 2013 #5
    I answered your question the best I could.

    Your question is closely related to the question I am posing. If I could simply answer it, then I wouldn't've started the thread.
  7. May 29, 2013 #6

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Try again, but this time answer the question. Start by answering this question: What is the complement (in [0,1]) of ##\cap_{k=1}^{\infty} E_k##? When you have answered this, can you find an upper bound on the measure of that complement?
  8. May 29, 2013 #7
    By DeMorgan, [itex]C\left(\cap_{k=1}^{\infty} E_k\right)=\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}[/itex]

    A lower bound immediately comes to mind. I'm still stewing on how to get an upper bound to pop out.

    Lower Bound:
    Consider: [itex]m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right][/itex]. [itex]m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]= m\left[0,1\right]-m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\geq m\left[0,1\right]-\Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=1-0=1[/itex]
  9. May 29, 2013 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You have a set which is a subset of [0,1] and has measure bigger than or equal to 1, and you're not sure what its measure is?
  10. May 29, 2013 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In Step 1 you did not follow through all the way: if ##E = \cap_k E_k,## then
    [tex] E^c = \bigcup_k E_k^c [/tex]
  11. May 30, 2013 #10
    Yep. I messed that up too. One more time.

    By DeMorgan, [itex]C\left(\cap_{k=1}^{\infty} E_k\right)=C\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]=\bigcup_{k=1}^{\infty}E^{c}_{k}[/itex]

    Consider: [itex]m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)[/itex]. [itex]m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\leq \Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=0[/itex]

    Therefore, [itex]m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)=0\Rightarrow m\left(\cap_{k=1}^{\infty}E_{k}\right)=1[/itex]
  12. May 30, 2013 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Easier: from ##E = \cap_k E_k## we have ##E^c = \cup_k E_k^c## (no need for all those intermediate steps). Then [tex] 0 \leq m(E^c) \leq \sum_k m(E_k^c) = \sum_k 0 =0,[/tex] where the ##`\leq '## is from some standard property or result, depending on exactly what approach was used in the course/notes/textbook.
    Last edited: May 30, 2013
  13. May 30, 2013 #12
    I like the sandwhich approach better than mine.

    Thanks for the pointers. I really lost the forest from the trees on this problem.

    I got it stuck in my head that perhaps the irrationals were "dense enough" to create 2 subsets with the same measure. It didn't dawn on me to use L.measure properties to disprove the existence of that set.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted